Jim, he put 4 LEDs in series, each with it's '2.8' volt drop, and had only the remaining voltage to drop across. It does
look incorrect, however, in the voltage to drop across. 4 x 2.8 is 11.2. While-running voltage in 12 volt system is 14.2
(to use your number) leaves 3 volts to drop across....

--
Dan D.



..
"Jim Weir" wrote in message ...
Before everybody in the Western Hemisphere blows a bucket full of light emitting
diodes, would you care to calculate the resistor one more time? And perhaps
post a retraction?


(Jay)
shared these priceless pearls of wisdom:

-You can find examples on how to power the LEDs on the manufacturer web
-site.
-
-Having said that...


So lets say the recommended current for
-the LED is 20mA. Ohms law is R=E/I, so that gives you a resistor
-value of .3V/.02A=15 ohms.


Um, no. Suppose the diode has a forward voltage drop of 2.8 volts (that's not a
common value, but I'll give it to you for argument.

Now the power supply (battery) is a 12 volt supply, but 14.2 volts at full
charge with the alternator going, so the drop across the series resistor is
going to be

14.2 minus 2.8 equals 11.4 volts, which is the voltage across the resistor.
This current limiting resistor is going to have 20 mA flowing through it, so Ohm
tells us that resistance equals voltage divided by current. In this case, 11.4
volts divided by 20 mA gives us a resistor of 570 ohms (560 is the nearest
standard value).

You put your calculated 15 ohm resistor in series with this diode and I
guarantee you that the SNAP you hear is the gallium aluminum arsenide
semiconductor of the diode being sacrificed on Ohm's altar.

I'm serious. You owe the newsgroup a correction before somebody takes your
error and blows up a whole bunch of LEDs.

Jim


Jim Weir (A&P/IA, CFI, & other good alphabet soup)
VP Eng RST Pres. Cyberchapter EAA Tech. Counselor
http://www.rst-engr.com